How to solve $p^2 - p + 1 = q^3$ over primes?
I checked that $p$ and $q$ must be odd and I don't know what to do next. I don't have any experience with such equations.
How to solve $p^2 - p + 1 = q^3$ over primes?
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elementary-number-theory
prime-numbers
diophantine-equations
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2Rewrite as $p(p-1)=q^3-1=(q-1)(q^2+q+1)$. – 2017-02-23
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2write your equation in the form $$p(p-1)=(q-1)(q^2+q+1)$$ – 2017-02-23
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1$p$ is a primitive sixth root of $1$ modulo $q$, so $q-1$ must be divisible by $6$ . – 2017-02-23
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0@vadim123 $p$ must divide $(q - 1)$ or $(q^2 + q + 1)$ (only one of these two, because $(p - 1)$ is not a prime and is smaller than $p$). I really can't see what's next. – 2017-02-23
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1Better to study the given problem in the ring of Eisenstein integers $\mathbb{Z}[\omega]$, that is a UFD. – 2017-02-23
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0It is also interesting to point out that the problem can be put in the following form $$ (4q)^3 - (8p-4)^2 = 48$$ and there probably are just a finite number of solutions of $a^3-b^2=48$. – 2017-02-23
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3More general https://artofproblemsolving.com/community/c6h35935p224998 – 2017-02-23